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a: Sửa đề: \(27x^3-27x^2+9x-1\)
\(=\left(3x-1\right)^3\)
b: \(=\left(x-1\right)^3\)
c: \(=\left(x+\dfrac{1}{3}\right)\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)
\(a,\left(-3x+2\right)^3\)
\(=\left(2-3x\right)^3\)
\(=2^3-3.2^2.3x+3.2.9x^2-27x^3\)
\(=8-36x+54x^2-27x^3\)
\(a,\left(-3x+2\right)^3=\left(2-3x\right)^3=2^3-3.2^2.3x+3.2.9x^2-27x^3=8-36x+54x^2-27x^3\)
x3-3x2+3x-1=x3-3x2.1+3x.12-13
=(x-1)3
thay x=101 ta được:
(101-1)3=1003=1000000
x3+9x2+27x+27=x3+3x2.3+3.x.32+33
=(x+3)3
thay x=97 ta được:
(97+3)3=1003=1000000
x^3-3x^2+3x-1
x^3+9x^2+27x+27
=x^3+3*x^2*3+3*x*3^2+3^3
=(x+3)^3
thayx=97 ta duoc :(97+3)^3=100^3=1000000
\(\left(a-1\right)^3-\left(a+1\right)^3=a^3-3a^2+3a-1-a^3-3a^2-3a-1=-6a^2-2=-2\left(3a^2+1\right)\)
\(\left(a-1\right)^3-\left(a+1\right)^3\)
\(=\left[\left(a-1\right)-\left(a+1\right)\right]\left[\left(a-1\right)^2+\left(a-1\right)\left(a+1\right)+\left(a+1\right)^2\right]\)
\(=\left(-2\right)\left(a^2-2a+1+a^2-1+a^2+2a+1\right)\)
\(=\left(-2\right)\left(3a^2+1\right)\)
Câu a : \(x^3y^3+225=\left(xy\right)^3+\left(\sqrt[3]{225}\right)^3=\left(xy+\sqrt[3]{225}\right)\left(x^2y^2-xy\sqrt[3]{225}+\left(\sqrt[3]{225}\right)^2\right)\)
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Câu a : ( Mình nghĩ đề sai )
Câu b : \(x^3-3x^2+3x-1=\left(x-1\right)^3=\left(x-1\right)\left(x-1\right)\left(x-1\right)\)
Câu c : \(\dfrac{1}{27}+x^3=\left(\dfrac{1}{3}\right)^3+x^3=\left(\dfrac{1}{3}+x\right)\left(\dfrac{1}{9}-\dfrac{1}{3}x+x^2\right)\)
Câu d : \(0,001-1000x^3=\left(\dfrac{1}{10}\right)^3-\left(10x\right)^3=\left(\dfrac{1}{10}-10x\right)\left(\dfrac{1}{100}+x+100x^2\right)\)
Chúc bạn học tốt